What is the largest value of non negative integer N for which 10^(N) is a factor of 50!?

1. Translate the problem requirements: We need to find the largest value of \(\mathrm{N}\) such that \(10^{\mathrm{N}}\) divides \(50!\) evenly. Since \(10 = 2 \times 5\), we need \(10^{\mathrm{N}} = 2^{\mathrm{N}} \times 5^{\mathrm{N}}\) to be a factor of \(50!\)
2. Identify the limiting factor: Since \(10^{\mathrm{N}}\) requires both \(2^{\mathrm{N}}\) and \(5^{\mathrm{N}}\) as factors, we need to count how many factors of 2 and 5 are in \(50!\), then determine which is the limiting constraint
3. Count factors of 5 in \(50!\): Count multiples of 5, 25, 125, etc. in the numbers 1 through 50, since factors of 5 are typically less abundant than factors of 2
4. Verify factor count and determine maximum N: The limiting factor (between 2s and 5s) determines the largest possible value of \(\mathrm{N}\)

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking. We want to find the largest value of \(\mathrm{N}\) such that \(10^{\mathrm{N}}\) divides \(50!\) evenly - meaning when we divide \(50!\) by \(10^{\mathrm{N}}\), we get a whole number with no remainder.

Now, what does \(10^{\mathrm{N}}\) actually mean? Let's think about this concretely:
• \(10^1 = 10 = 2 \times 5\)
• \(10^2 = 100 = 2^2 \times 5^2\)
• \(10^3 = 1000 = 2^3 \times 5^3\)

So in general, \(10^{\mathrm{N}} = 2^{\mathrm{N}} \times 5^{\mathrm{N}}\). This means for \(10^{\mathrm{N}}\) to be a factor of \(50!\), we need both \(2^{\mathrm{N}}\) and \(5^{\mathrm{N}}\) to be factors of \(50!\).

Process Skill: TRANSLATE - Converting the abstract question about factors into concrete requirements for counting prime factors

2. Identify the limiting factor

Here's the key insight: since \(10^{\mathrm{N}}\) requires both \(2^{\mathrm{N}}\) and \(5^{\mathrm{N}}\) as factors, we're limited by whichever prime factor appears less frequently in \(50!\).

Think of it like making pairs of socks - if you have 20 left socks and 12 right socks, you can only make 12 complete pairs. Similarly, if \(50!\) contains many factors of 2 but fewer factors of 5, then the number of factors of 5 will limit how large \(\mathrm{N}\) can be.

In factorials, there are generally more factors of 2 than factors of 5, because every even number contributes at least one factor of 2, but only every 5th number contributes a factor of 5. So we expect that counting factors of 5 will give us our answer.

3. Count factors of 5 in \(50!\)

Now let's count how many factors of 5 appear in \(50!\). Remember, \(50! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times \ldots \times 50\).

We need to count factors of 5 from all numbers 1 through 50:

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
That's 10 numbers, so we get 10 factors of 5.

But wait! Some numbers contribute more than one factor of 5:
• Multiples of 25 (which is \(5^2\)) contribute an additional factor of 5
• In our range: 25, 50
• That's 2 additional factors of 5

Multiples of 125: Since 125 > 50, there are no multiples of 125 in our range.

Total factors of 5: 10 + 2 = 12

Let's verify this makes sense by listing some key contributions:
• \(25 = 5^2\) contributes 2 factors of 5
• \(50 = 2 \times 5^2\) contributes 2 factors of 5
• Each other multiple of 5 contributes exactly 1 factor of 5

Process Skill: INFER - Recognizing that some numbers contribute multiple factors of the same prime

4. Verify factor count and determine maximum N

We found that \(50!\) contains exactly 12 factors of 5.

Since factors of 2 are much more abundant in factorials (every even number contributes at least one factor of 2), we can be confident that \(50!\) contains at least 12 factors of 2. Therefore, the limiting factor is indeed the number of 5s.

This means the largest value of \(\mathrm{N}\) for which \(10^{\mathrm{N}} = 2^{\mathrm{N}} \times 5^{\mathrm{N}}\) divides \(50!\) is \(\mathrm{N} = 12\).

Let's double-check: \(10^{12} = 2^{12} \times 5^{12}\). Since \(50!\) contains at least 12 factors of 2 and exactly 12 factors of 5, we can form 12 complete factors of 10. We cannot form a 13th factor of 10 because we would need 13 factors of 5, but we only have 12.

4. Final Answer

The largest value of \(\mathrm{N}\) for which \(10^{\mathrm{N}}\) is a factor of \(50!\) is \(\mathrm{N} = 12\).

The answer is C. 12.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "factor" means
Students may confuse "\(10^{\mathrm{N}}\) is a factor of \(50!\)" with "\(50!\) is a factor of \(10^{\mathrm{N}}\)" or think it means finding how many times 10 appears in the multiplication \(1\times 2\times 3\times \ldots\times 50\), rather than understanding that we need \(10^{\mathrm{N}}\) to divide \(50!\) evenly.

2. Not recognizing the prime factorization connection
Students might try to directly count occurrences of "10" in the factorial expansion instead of realizing that \(10^{\mathrm{N}} = 2^{\mathrm{N}} \times 5^{\mathrm{N}}\), so they need to count prime factors of 2 and 5 separately.

3. Forgetting about the limiting factor principle
Students may attempt to count both factors of 2 and factors of 5 separately and then add them together, not understanding that since we need BOTH \(2^{\mathrm{N}}\) and \(5^{\mathrm{N}}\), we're limited by whichever appears less frequently.

Errors while executing the approach

1. Missing multiples that contribute extra factors
The most common error is counting each multiple of 5 only once. Students often count 25 and 50 as contributing just one factor of 5 each, missing that \(25 = 5^2\) contributes 2 factors and \(50 = 2\times 5^2\) contributes 2 factors of 5.

2. Arithmetic errors in systematic counting
Students may miscount the multiples of 5 (getting 9 or 11 instead of 10) or make errors when applying the formula: \(\lfloor 50/5 \rfloor + \lfloor 50/25 \rfloor + \lfloor 50/125 \rfloor + \ldots\), especially miscalculating \(\lfloor 50/25 \rfloor = 2\).

3. Incorrectly handling the powers of 5
Some students might double-count or under-count by incorrectly thinking that 25 contributes 3 factors (confusing \(5^2 = 25\) with \(5^3\)), or forgetting that every multiple of 25 is also a multiple of 5.

Errors while selecting the answer

1. Selecting the count of multiples instead of total factors
Students who correctly identify that there are 10 multiples of 5 from 1 to 50 might select answer choice A (5) or mistakenly choose an answer related to just counting the multiples rather than the total factor count of 12.